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Advanced Calculus for Engineers

This document summarizes key concepts about partial derivatives of multivariable functions: 1) It defines partial derivatives as differentiating a function of more than one variable with respect to a single variable, while holding other variables constant. Notation and formulas for first and second order partial derivatives are provided. 2) An example calculates the first order partial derivatives fx and fy for a sample function f(x,y) = xex3y + 4x - y2 - 4x2y4 at the point (1, ln3). 3) A second example finds the second order partial derivatives fxx, fyy, fxy, and fyx for the function f(x,y) =

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Muhammad Nasrul
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38 views5 pages

Advanced Calculus for Engineers

This document summarizes key concepts about partial derivatives of multivariable functions: 1) It defines partial derivatives as differentiating a function of more than one variable with respect to a single variable, while holding other variables constant. Notation and formulas for first and second order partial derivatives are provided. 2) An example calculates the first order partial derivatives fx and fy for a sample function f(x,y) = xex3y + 4x - y2 - 4x2y4 at the point (1, ln3). 3) A second example finds the second order partial derivatives fxx, fyy, fxy, and fyx for the function f(x,y) =

Uploaded by

Muhammad Nasrul
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Civil Engineering Mathematics III

Chapter 1
Multivariable Function
1.3.1 DEFINITION OF PARTIAL DERIVATIVES
1.3.2 MIXED DERIVATIVES
1.3.1 Partial Derivatives
𝑑𝑦
Previously, differentiate for single variable - or 𝑦 ′
𝑑𝑥
𝜕 𝜕
Currently, differentiate for more than one variable or
𝜕𝑥 𝜕𝑦

Notation
Let 𝑧 = 𝑓 𝑥, 𝑦
𝜕𝑓(𝑥,𝑦) 𝜕𝑓 𝜕𝑧
Partial derivatives with respect to 𝑥 - 𝑓𝑥 , 𝑓𝑥 𝑥, 𝑦 , , , , 𝑧𝑥
𝜕𝑥 𝜕𝑥 𝜕𝑥
𝜕𝑓(𝑥,𝑦) 𝜕𝑓 𝜕𝑧
Partial derivatives with respect to 𝑦 - 𝑓𝑦 , 𝑓𝑦 𝑥, 𝑦 , , , , 𝑧𝑦
𝜕𝑦 𝜕𝑦 𝜕𝑦

𝑓 𝑥+ℎ,𝑦 −𝑓(𝑥,𝑦)
First Partial Derivatives 𝑓𝑥 𝑥, 𝑦 = lim use this formula or convenient way
ℎ→0 ℎ
Example 1: 1st partial derivatives
𝟑𝒚
Given 𝒇 𝒙, 𝒚 = 𝒙𝒆𝒙 + 𝟒𝒙 − 𝒚𝟐 − 𝟒𝒙𝟐 𝒚𝟒 , find 𝑓𝑥 and 𝑓𝑦 . Evaluate each at the point (1, ln 3)
Step 1: 𝑓𝑥 , 𝑥 only variable, other constant
3𝑦 3𝑦 3𝑦 3𝑦
𝑓𝑥 = 𝑥 3𝑥 2 𝑦 𝑒 𝑥 + 𝑒𝑥 + 4 − 0 − 8𝑥𝑦 4 = 3𝑥 3 𝑦𝑒 𝑥 + 𝑒𝑥 + 4 − 8𝑥𝑦 4
𝑓𝑥 (1, ln 3) = 3 1 3 (ln 3) 𝑒 1 3 (𝑙𝑛3) +𝑒 1 3 (ln 3) + 4 − 8(1)(ln 3)4 = 9ln 3 + 7 − 8(ln 3)4
Step 2: 𝑓𝑦 , 𝑦 only variable, other constant
3𝑦 3𝑦
𝑓𝑦 = 𝑥 𝑥 3 𝑒 𝑥 + 0 − 2𝑦 − 16𝑥 2 𝑦 3 = 𝑥 4 𝑒 𝑥 − 2𝑦 − 16𝑥 2 𝑦 3
3
𝑓𝑦 (1, ln 3) = 14 𝑒 1 ln 3
− 2 ln 3 − 1612 ln 33 = 3 − 2 ln 3 − 16 (ln 3)3 𝑦 = 𝑢𝑣
𝑑𝑦 𝑑𝑣 𝑑𝑢
=𝑢 +𝑣
𝑑𝑥 𝑑𝑥 𝑑𝑥
Second-Order Partial Derivatives
𝜕 𝜕𝑓 𝜕2 𝑓
= = 𝑓𝑥𝑥
𝜕𝑥 𝜕𝑥 𝜕𝑥𝜕𝑥
𝜕 𝜕𝑓 𝜕2 𝑓
= = 𝑓𝑦𝑦
𝜕𝑦 𝜕𝑦 𝜕𝑦𝜕𝑦
𝜕 𝜕𝑓 𝜕2 𝑓
= = 𝑓𝑥𝑦
𝜕𝑦 𝜕𝑥 𝜕𝑥𝜕𝑦
𝜕 𝜕𝑓 𝜕2 𝑓
= = 𝑓𝑦𝑥
𝜕𝑥 𝜕𝑦 𝜕𝑦𝜕𝑥

𝑓𝑥𝑦 = 𝑓𝑦𝑥
Example 2: 2nd order partial derivatives
Find 𝑓𝑥𝑥 , 𝑓𝑦𝑦 , 𝑓𝑥𝑦 and 𝑓𝑦𝑥 for 𝒇 𝒙, 𝒚 = 𝒙(𝟐𝒙 + 𝒚) −𝟏 + 𝐬𝐢𝐧 𝒙𝒚 y = sin 𝑎𝑥
𝑦 ′ = 𝑎 cos 𝑥
−2
𝑓𝑥 = 𝑥 −1 2𝑥 + 𝑦 2 + (2𝑥 + 𝑦)−1 +𝑦 cos 𝑥𝑦
= −2x 2𝑥 + 𝑦 −2 + 2𝑥 + 𝑦 −1 + 𝑦 cos 𝑥𝑦
𝑓𝑥𝑥 = −2 2𝑥 + 𝑦 −2 − 2𝑥 −2 2𝑥 + 𝑦 −3 2 − 2𝑥 + 𝑦 −2 2 − 𝑦 2 sin 𝑥𝑦
−4 8𝑥
= + − 𝑦 2 sin 𝑥𝑦
(2𝑥+𝑦)2 (2𝑥+𝑦)3

𝑓𝑥𝑦 = −2𝑥 −2 2𝑥 + 𝑦 −3 1 − 2𝑥 + 𝑦 −2 1 − 𝑦[𝑥 sin 𝑥𝑦] + cos 𝑥𝑦


4𝑥 1
= − − 𝑥𝑦 sin 𝑥𝑦 + cos 𝑥𝑦
(2𝑥+𝑦)3 2𝑥+𝑦 2

Find yourself 𝑓𝑦 , 𝑓𝑦𝑦 and 𝑓𝑦𝑥 !!

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